5.10 Linear Approximations
In general from the equation of the tangent plane above, an equation of the tangent at the point \((a,b,f(a,b))\) is \[z = f(a,b) + f_x(a,b)(x -a) + f_y(a,b)(y - b)\] The linear function whose graph is the tangent plane, namely: \[L(x,y) = f(a,b) + f_x(a,b)(x -a) + f_y(a,b)(y - b)\] is called the linearization of \(f\) at \((a,b)\) and the approximation. \[f(x,y) \approx z = f(a,b) + f_x(a,b)(x -a) + f_y(a,b)(y - b)\] is called the standard linear approximation or the tangent plane approximation of \(f\) at \((a,b)\).
- 1.
- Given the function \(\displaystyle {f(x,y) = xe^{xy}}\). Find its linearization at \((1,0)\). Then use it to approximate \(f(1.1,-0.1)\).
- 2.
- Find the linearization of \(f(x,y,z) = \sqrt {x^2 + y^2 + z^2}\) at \((3,2,6)\) and use it to approximate the value of \(\sqrt {(3.02)^2 + (1.97)^2 + (5.99)^2}\)
Solution.
Part 1
\(f(a,b) = f(1,0) = 1 \hspace {0.2cm} , \hspace {0.2cm} f_x(x,y) = \displaystyle {e^{xy} + xye^{xy}}\hspace {0.2cm} , \hspace {0.2cm} f_y(x,y) = \displaystyle {x^2 e^{xy}}\)
\(f_x(1,0) = 1\hspace {0.2cm} , \hspace {0.2cm} f_y (1,0) = 1\). So the linearization of \(f\) at \((1,0)\) is \begin {align*} L(x,y) & = 1 + 1(x-1) + 1(y-0)\\ & = 1 + x - 1 + y\\ & = x+y\\\\ \therefore \hspace {0.4cm} L(x,y) & = x + y \end {align*}
\begin {align*} f(x,y) & \approx L(x,y) = x + y\\ f(1.1,0.1) & \approx 1.1 + (-0.1)\\ & = 1.1 - 0.1\\ & = 1 \end {align*}
Compare with actual value of \(f(1.1,-0.1) = \displaystyle {1.1e^{-(1.1)(0.1)}}\approx 0.98542\)
Part 2
\[L(x,y,z) = f(3,2,6) + f_x(3,2,6)(x-3) + f_y(3,2,6)(y-2) + f_z(3,2,6)(z - 6)\]
\(f(3,2,6) = \sqrt {9 + 4 + 36} = \sqrt {49}= 7\)
\begin {align*} f_x(x,y,z) & = \dfrac {1}{2}\Big ( x^2 + y^2 + z^2\Big )^{-\dfrac {1}{2}}\cdot 2x\\ & = \dfrac {x}{\sqrt {x^2 + y^2 + z^2}}\\ & = \dfrac {x}{f(x,y,z)} \end {align*}
Similarly, \(\displaystyle {f_y(x,y,z) = \dfrac {y}{f(x,y,z)}}\hspace {0.2cm} ,\hspace {0.2cm} \displaystyle {f_z(x,y,z) = \dfrac {z}{f(x,y,z)}}\)
\(f_x(3,2,6) = \dfrac {3}{f(3,2,6)} = \dfrac {3}{7}\)
\(f_y(3,2,6) = \dfrac {2}{7}\)
\(f_z(3,2,6) = \dfrac {6}{7}\)
\begin {align*} L(x,y,z) & = 7 + \dfrac {3}{7}(x - 3) + \dfrac {2}{7}(y - 2) + \dfrac {6}{7}(z - 6)\\ & = \dfrac {3}{7}x + \dfrac {2}{7}y + \dfrac {6}{7}z + 7 - \frac {9}{7} - \frac {4}{7} - \frac {36}{7}\\\\ \therefore \hspace {0.2cm} L(x,y,z) & = \dfrac {1}{7}\big (3x + 2y + 6z\big ) \end {align*}
\[f(x,y,z) \approx L(x,y,z) = \dfrac {1}{7}\big (3x + 2y + 6z\big )\]
We seek \(\displaystyle {f(3.02,1.97,5.99) = \dfrac {1}{7}\Big [3(3.02) + 2(1.97) + 6(5.99)\Big ]}\) so that \begin {align*} \sqrt {(3.02)^2 + (1.97)^2 + (5.99)^2} & \approx \dfrac {1}{7}\Big (9.06 + 3.94 + 35.94\Big )\\ & = \dfrac {1}{7}\times 48.94\\ & = 6.99\\\\ \end {align*}
The Error in the Standard Linear Approximation of \(f(x,y)\) Near \((a,b)\)
If \(f\) and its second partial derivative are continuous throughout a rectangle \(R\) centred at \((a,b)\), then the error \(E(x,y)\)
incurred in replacing \(f(x,y)\) on \(R\) by its linearization.
\[L(x,y,z) = f(a,b) + f_x (a,b)(x - a) + f_y (a,b)(x - b) \]
satisfies the inequality
\[\left |E(x,y)\right | \leq \dfrac {1}{2}M\Big [\left |x - a\right | + \left |y - a\right |\Big ]^2\]
where \(M\) is any upper bound for the values \(\left |f_{xx}\right |\hspace {0.2cm} , \hspace {0.2cm} \left |f_{xy}\right |\) and \(\left |f_{yy}\right |\) on \(R\).
Find the linearization of \(f(x,y) = x^2 - xy + \dfrac {1}{2}y^2 + 3\hspace {0.2cm}\) near \((3,2)\). Hence find an upper bound for the error in the approximation \(f(x,y) \approx L(x,y)\) over the rectangle \[R:\hspace {0.3cm} \left |x - 3\right | \leq 0.1\hspace {0.3cm} , \hspace {0.3cm} \left |y - 2\right |\leq 0.1\]
Solution.
\(f(3,2) = 9 - 6 + 2 + 3 = 8\)
\(f_{x}(x,y) = 2x- y \implies f_x(3,2) = 4\)
\(f_y(x,y) = -x + y \implies f_y(3,2) = -1\)
\begin {align*} L(x,y) & = f(3,2) + f_x(3,2)(x-3) + f_y(3,2)(y-2)\\ & = 8 + 4(x-3) -(y - 2)\\ & = 4x -y -2 \end {align*}
\(f_{xx} = 2 \hspace {0.2cm} , \hspace {0.2cm} f_{xy} = -1 \hspace {0.2cm} , \hspace {0.2cm} f_{yy}= 1\). The largest of these is ? So \(M = 2\)
\[\left |E\right |\leq \dfrac {1}{2}M\Big [\left |x - 3\right | + \left |y - 2\right |\Big ]^2\]
But \(\left |x - 3\right |\leq 0.1\) and \(\left |y - 2\right |\leq 0.1\)
\[\left |E\right |\dfrac {1}{2}M\Big [\left |x - 3\right | + \left |y - 2\right |\Big ]^2 = \Big [(0.1)^2 + (0.1)^2\Big ]^2 - (0.2)^2 = 0.04\]
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